Setze ${\displaystyle F_{n,k}(x,y)={\tbinom {n}{k}}(x+k)^{k-1}\,(y-k)^{n-k}}$,

${\displaystyle G_{n,k}(x,y)=(y-n){\tbinom {n-1}{k-1}}(x+k)^{k-1}\,(y-k)^{n-k-1}}$ und ${\displaystyle S_{n}(x,y)=\sum _{k=0}^{n}F_{n,k}(x,y)}$.

Es ist ${\displaystyle F_{n,k}(x,y)-yF_{n-1,k}(x,y)=(y-k){\tbinom {n}{k}}(x+k)^{k-1}\,(y-k)^{n-k-1}-y{\tbinom {n-1}{k}}\,(x+k)^{k-1}\,(y-k)^{n-1-k}}$

${\displaystyle {\Big (}y\left({\tbinom {n}{k}}-{\tbinom {n-1}{k}}\right)-k{\tbinom {n}{k}}{\Big )}\,(x+k)^{k-1}\,(y-k)^{n-1-k}=\left(y{\tbinom {n-1}{k-1}}-n{\tbinom {n-1}{k-1}}\right)\,(x+k)^{k-1}\,(y-k)^{n-1-k}=G_{n,k}(x,y)}$.

und ${\displaystyle (x+n)F_{n-1,k}(x+1,y-1)-G_{n,k+1}(x,y)\,}$

${\displaystyle =(x+n)(y-1-k){\tbinom {n-1}{k}}(x+1+k)^{k-1}\,(y-1-k)^{n-k-2}-(y-n)(x+k-1){\tbinom {n-1}{k}}(x+k+1)^{k-1}(y-k-1)^{n-k-2}}$

${\displaystyle ={\Big (}(x+n)(y-1-k)-(y-n)(x+k+1){\Big )}{\tbinom {n-1}{k}}(x+1+k)^{k-1}(y-1-k)^{n-k-2}}$

${\displaystyle =(x+y)(n-1-k){\tbinom {n-1}{k}}(x+1+k)^{k-1}(y-1-k)^{n-k-2}=(x+y)(n-1){\tbinom {n-2}{k}}(x+1+k)^{k-1}(y-1-k)^{n-k-2}}$

${\displaystyle =(x+y)(n-1)F_{n-2,k}(x+1,y-1)\,}$.

Also gilt

${\displaystyle {\text{(i)}}\;\,\quad F_{n,k}(x,y)-yF_{n-1,k}(x,y)=G_{n,k}(x,y)}$

${\displaystyle {\text{(ii)}}\quad (x+n)F_{n-1,k}(x+1,y-1)-(x+y)(n-1)F_{n-2,k}(x+1,y-1)=G_{n,k+1}(x,y)}$

Summiert man die Differenz ${\displaystyle {\text{(i)}}-{\text{(ii)}}\,}$

${\displaystyle F_{n,k}(x,y)-yF_{n-1,k}(x,y)-(x+n)F_{n-1,k}(x+1,y-1)+(n-1)(x+y)F_{n-2,k}(x+1,y-1)=G_{n,k}(x,y)-G_{n,k+1}(x,y)\,}$

nach ${\displaystyle k\,}$ von ${\displaystyle 0\,}$ bis ${\displaystyle n\,}$, so ist ${\displaystyle S_{n}(x,y)-yS_{n-1}(x,y)-(x+n)S_{n-1}(x+1,y-1)+(n-1)(x+y)S_{n-2}(x+1,y-1)=0\,}$,

da das Teleskop-Produkt ${\displaystyle \sum _{k=0}^{n}{\Big (}G_{n,k}(x,y)-G_{n,k+1}(x,y){\Big )}=G_{n,0}(x,y)-G_{n,n+1}(x,y)}$ verschwindet.

Für ${\displaystyle n=0,1\,}$ gilt ${\displaystyle S_{n}(x,y)={\frac {(x+y)^{n}}{x}}}$. Der Induktionsschluss von ${\displaystyle n-2\,}$ und ${\displaystyle n-1\,}$ auf ${\displaystyle n\,}$ ergibt sich aus der Rekursionsformel:

${\displaystyle S_{n}(x,y)-y{\frac {(x+y)^{n-1}}{x}}-(x+n)\,{\frac {(x+y)^{n-1}}{x+1}}+(n-1)(x+y){\frac {(x+y)^{n-2}}{x+1}}=0\Rightarrow S_{n}(x,y)={\frac {(x+y)^{n}}{x}}}$

In der Formel ${\displaystyle \sum _{k=0}^{n}{n \choose k}(x+k)^{k-1}\,(y-k)^{n-k}={\frac {(x+y)^{n}}{x}}}$ ersetze ${\displaystyle (x,y)\,}$ durch ${\displaystyle \left({\frac {x}{a}},{\frac {y}{a}}\right)}$ und multipliziere die so

entstandene Gleichung ${\displaystyle \sum _{k=0}^{n}{n \choose k}\left({\frac {x}{a}}+k\right)^{k-1}\,\left({\frac {y}{a}}-k\right)^{n-k}={\frac {1}{a^{n-1}}}{\frac {(x+y)^{n}}{x}}}$ mit ${\displaystyle x\,a^{n-1}}$ durch.

Dann ist ${\displaystyle \sum _{k=0}^{n}{n \choose k}x\,(x+ak)^{k-1}\,(y-ak)^{n-k}=(x+y)^{n}}$.